Posted by Christina on Monday, March 18, 2013 at 3:27pm.
The mean number of bumped airline passengers per day is 1.35 and the standard deviation is 0.25. For a random selection of a group of 40 days, what is the probability that the mean of bumped passengers for the 40 days will be between 1.25 and 1.50?
stats. - MathGuru, Monday, March 18, 2013 at 6:02pm
Use this z-score formula for this problem:
z = (x - mean)/(sd/√n)
x = 1.25, 1.50
mean = 1.35
sd = 0.25
n = 40
Calculate two z-scores, then use a z-table to determine probability between the two scores.
I hope this will help get you started.
stats. - Christina, Monday, March 18, 2013 at 9:57pm
but not sure how to read the z scores I don't know if it would be.5000-.5000 or what can someone please help me on the z scores for -0.0791 and 0.0791
stats. - PsyDAG, Tuesday, March 19, 2013 at 3:01pm
SD/√n = .25/6.3246 = .039528192 = .04
Z = (1.25-1.35)/.04 = -.10/.04 = -2.5
Z = (1.50-1.35)/.04 = .15/.04 = ?
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability of both Z scores between those Z scores and mean. Add the two together.
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